# The Unapologetic Mathematician

## Calculating the Degree of a Proper Map

As I asserted yesterday, there is a simple formula for the degree of a proper map $f:M\to N$. Pick any regular value $q\in N$ of $f$, some of which must exist, since the critical values have measure zero in $N$. For each $p\in f^{-1}(q)$ we define $\mathrm{sgn}_pf$ to be $+1$ or $-1$ as $f$ is orientation-preserving or orientation-reversing. Then I say that $\displaystyle\deg f=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf$

The key is the inverse function theorem: the Jacobian of $f$ must have maximal rank at $p$, so there’s some $U\subseteq M$ around $p$ on which $f:U\to f(U)\subseteq N$ is a diffeomorphism. This is true for each $p\in f^{-1}(q)$, so we get an open $U_p$ around each $p$. Since $f$ is proper, there are only finitely many such $p$, and they’re all separated from each other, so we can shrink each $U_p$ so that they’re all disjoint. Then we can set $V\subseteq N$ to be the intersection of all the $f(U_p)$, and we can further shrink each $U_p$ until its image is exactly $V$.

That is, we’ve started with a regular value $q\in N$ and its preimage $f^{-1}(q)$, consisting of a bunch of points $p\in M$; we’ve used the fact that $f$ is proper to widen it to an open set $V$ and its preimage, consisting of a bunch of disjoint open sets $U_p\subseteq M$. Since $f:U_p\to V$ is a diffeomorphism for each $p\in f^{-1}(q)$, $f$ is constantly orientation-preserving or orientation-reversing over all of $U_p$, though of course each $U_p$ can be different.

Now if we pick some $\omega\in\Omega_c^n(N)$ supported in $V$, then $f^*\omega$ is supported in the disjoint union of the $U_p$. So we can calculate \displaystyle\begin{aligned}\int\limits_Mf^*\omega&=\int\limits_{\biguplus_{p\in f^{-1}(q)}U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\int\limits_{U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\int\limits_{f(U_p)}\omega\\&=\left(\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\right)\int\limits_V\omega\\&=\deg f\int\limits_V\omega\end{aligned}

Since the coefficient $\deg f$ is independent of the choice of $\omega$, this proves the assertion.

December 10, 2011 - Posted by | Differential Topology, Topology

## 1 Comment »

1. […] what does this mean? The identity map has degree , while we can calculate that the antipodal map has degree . Since these are different, the two maps must not act […]

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